If $$X = \left\{ {{4^n} - 3n - 1:n \in N} \right\}$$ and $$Y = \left\{ {9\left( {n - 1} \right):n \in N} \right\},$$ where $$N$$ is the set of natural numbers, then $$X \cup Y$$ is equal to:
A.
$$X$$
B.
$$Y$$
C.
$$N$$
D.
$$Y - X$$
Answer :
$$Y$$
Solution :
$$\eqalign{
& {4^n} - 3n - 1 \cr
& = {\left( {1 + 3} \right)^n} - 3n - 1 \cr
& = \left[ {^n{C_0} + {\,^n}{C_1}.3 + {\,^n}{C_2}{{.3}^2} + ..... + {\,^n}{C_n}{{.3}^n}} \right] - 3n - 1 \cr
& = 9\left[ {^n{C_2} + {\,^n}{C_3}.3 + ..... + {\,^n}{C_n}{{.3}^{n - 2}}} \right] \cr} $$
$$\therefore \,\,{4^n} - 3n - 1$$ is a multiple of 9 for all $$n.$$
∴ $$X$$ = {$$x : x$$ is a multiple of 9 }
Also, $$Y = \left\{ {9\left( {n - 1} \right):n \in N} \right\} = $$ {All multiples of 9}
Clearly $$X \subset Y.$$
$$\therefore X \cup Y = Y.$$
Releted MCQ Question on Calculus >> Sets and Relations
Releted Question 1
If $$X$$ and $$Y$$ are two sets, then $$X \cap {\left( {X \cup Y} \right)^c}$$ equals.